If $x \diamond y = 2x^{2}-y^{2}$ and $x \otimes y = 4y+4$, find $(-1 \diamond 4) \otimes 3$.
Solution: We don't need to find $-1 \diamond 4$ because $x \otimes y$ depends only on the right operand. Find $x \otimes 3$ $ x \otimes 3 = (4)(3)+4$ $ \hphantom{x \otimes 3} = 16$.